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78 votes
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What does it mean for a set to have Lebesgue measure zero?

Let me quote von Neumann twice: There's no sense in being precise when you don't even know what you're talking about.   Young man, in mathematics you don't understand things. You just get ...
Asaf Karagila's user avatar
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38 votes
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Are most matrices diagonalizable?

Yes. Here is a proof over $\mathbb{C} $. Matrices with repeated eigenvalues are cut out as the zero locus of the discriminant of the characteristic polynomial, thus are algebraic sets. Some matrices ...
AnonymousCoward's user avatar
37 votes
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Apparent inconsistency of Lebesgue measure

"As the rationals are dense in $\mathbb{R}$ we must have $\mathbb{R}\subset \bigcup_{n=1}^\infty I_n$." This is false. Pick your favorite irrational number $x$. For every $n$, there exist infinitely ...
Alex Kruckman's user avatar
36 votes

Intuitive, possibly graphical explanation of why rationals have zero Lebesgue measure

You could utilize one of the well known ways to count the rational numbers, namely consider the integer lattice $\mathbb Z^2$ and the subset $\{(a,b)\mid a\geq 1\ \wedge\ b\geq 0\}$ as illustrated ...
String's user avatar
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34 votes
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Intuitive, possibly graphical explanation of why rationals have zero Lebesgue measure

This is a really hard question; I think in general intuition for this sort of thing tends to come with experience, as you get used to the concepts. Having said that, I'll try to articulate the way ...
Sebastian Monnet's user avatar
30 votes

What does it mean for a set to have Lebesgue measure zero?

Practically speaking this means that when you integrate functions such sets don't matter, in the sense that if you modify the domain of integration by a set of measure zero, the integral is unchanged.
Mikhail Katz's user avatar
29 votes

Intuitive, possibly graphical explanation of why rationals have zero Lebesgue measure

This isn't a geometric answer, but you can get a lot of intuition for Lebesgue measure by thinking about it probabilistically. Specifically: The measure of a subset $S\subseteq [0,1)$ is the same ...
Jim Belk's user avatar
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27 votes

Is the measure of the sum equal to the sum of the measures?

Another example: $$A=\mathbb{Z},\quad B=[0,1],\quad A+B=\mathbb{R}. $$
Julián Aguirre's user avatar
27 votes

Prove that the graph of a measurable function is measurable.

Hint: Show that the mapping $$(x,y) \mapsto T(x,y) := f(x)-y$$ is measurable. Conclude that $\Gamma(f) = T^{-1}(\{0\})$ is measurable.
saz's user avatar
  • 121k
26 votes

A possible paradox about topology and the relation between zero and infinity?

There are a couple of different issues going on here: if we increase the number of dimension ... the volume increases Here's the first problem: the volume of an $n$-dimensional cube whose side ...
Billy's user avatar
  • 5,252
23 votes

Measure Theory - Why doesn't empty interior imply zero measure?

You have no reason to assume that the measure of the boundary is $0$ too. And you provided an example yourself: the irrationals. Its boundary is $\Bbb R$, whose Lebesgue measure is infinity.
José Carlos Santos's user avatar
22 votes
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Is diameter of a set a measure?

Observe that the diameter of singletons is $0$ and the diameter of set $\{x,y\}$ is $d(x,y)>0$ if $x\neq y$. So there is no additivity.
drhab's user avatar
  • 151k
21 votes
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The Lebesgue measure of zero set of a polynomial function is zero

Suppose the theorem is established for polynomials in $n-1$ variables. Let $p$ be a nontrivial polynomial in $n$ variables, say of degree $k \ge 1$ in $x_n$. We can then write $$p(\mathbf{x}, x_n) = ...
Nate Eldredge's user avatar
20 votes

Does a set with strictly positive Lebesgue measure contain an interval?

The irrationals are an easy example of not having this property.
J.G's user avatar
  • 5,094
19 votes
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Does a set with strictly positive Lebesgue measure contain an interval?

Nope. There is a standard construction of "fat Cantor sets" which have positive measure but are in fact nowhere dense, which is considerably stronger than merely containing no intervals. The ...
Ian's user avatar
  • 102k
19 votes

If $E$ has Lebesgue measure $0$, must there exist a translate such that $E\cap E+x=\varnothing$?

Let $C$ be the usual Cantor set: a measure zero subset of $[0,1]$. Let $C_\infty=\bigcup_{k=-\infty}^\infty(C+k)$, that is the union of the translates of $C$ by integers. Also $C_\infty$ has measure ...
Angina Seng's user avatar
18 votes
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What is wrong in this proof: That $\mathbb{R}$ has measure zero

Here is a slight variation on your construction. Consider $\mathbb{Q}$ which is countable, we may enumerate $\mathbb{Q}=\{q_1, q_2, \dots\}$. For each rational number $q_k$, cover it by an open ...
bof's user avatar
  • 78.8k
18 votes
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Are the measurable spaces $(\mathbb{R}^n, Bor(\mathbb{R}^n))$ and $(\mathbb{R}^m, Bor(\mathbb{R}^m))$ isomorphic for $n\neq m$

All the measure spaces $(\mathbb{R}^n, Bor(\mathbb{R}^n), \mu^n)$ are isomorphic. One quick way to prove this is to observe that $(\mathbb{R}, Bor(\mathbb{R}), \mu)$ is isomorphic to the product ...
Eric Wofsey's user avatar
18 votes

Measure Theory - Why doesn't empty interior imply zero measure?

As your example with irrational numbers shows, $\text{Bd}$ is not necessary $(n - 1)$-dimensional, it can very well be the entire space. (when talking about dimension for objects other than vector ...
mihaild's user avatar
  • 15.7k
17 votes

Is diameter of a set a measure?

It's not even finitely additive. If $X$ and $Y$ are two disjoint closed intervals on the real line then the diameter of their union is not the sum of their diameters.
Ethan Bolker's user avatar
17 votes
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Is the measure of the sum equal to the sum of the measures?

I preassume that here $A+B:=\{a+b\mid a\in A, b\in B\}$. Counterexample (for Lebesgue measure): Take $A=[0,1]\cup[2,3]$ and $B=[0,1]$ (so that $A+B=[0,4]$)
drhab's user avatar
  • 151k
17 votes
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What is the measure of the set of numbers in $[0,1]$ whose decimal expansions do not contain $5$?

See, study the complement of the set. That is, look at integers which contain $5$ in their decimal expansion. Caveat : note that $0.6 = 0.5\overline{9}$ also counts as a decimal which is expressed ...
Sarvesh Ravichandran Iyer's user avatar
16 votes
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Intuitive Explanation of Why the Power Set of $\mathbb{R}$ is "too big" for the Lebesgue Measure?

Let me try to explain the Vitali construction by breaking everything into small pieces. (Usually I find this presented quite rapidly in books, so that everything seems to come all at once and it's not ...
Ian's user avatar
  • 102k
16 votes

What is wrong in this proof: That $\mathbb{R}$ has measure zero

This part Each real number is arbitrarily close to a rational number since $\Bbb Q$ is dense in $\Bbb R$. Thus, each real number is in one of the open intervals. is not true. For an arbitrary $x\...
BigbearZzz's user avatar
  • 15.2k
16 votes
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Why "countability" in definition of Lebesgue measures?

If $\sum_{I \in X} l(I)=\infty$, then there is a countable subset of $X$ that sum up to infinity too. If $\sum_{I \in X} l(I)<\infty$, then only countably many terms of $l(I)$ is non-zero. In ...
BigbearZzz's user avatar
  • 15.2k
16 votes
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Counter example for $\int_\mathbb{R} f_nd\mu\rightarrow\int_\mathbb{R} fd\mu$

Consider the "melting ice cube" example (I like to have little mnemonics for remembering useful counterexmples): let $f_n := \frac{1}{n} \chi_{[0,n]}$. Then $f_n$ converges uniformly to zero, but $\...
Xander Henderson's user avatar
  • 30.1k
16 votes
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Integrating the following function

By Fubini, \begin{align*} I&=\int_{{\bf{R}}^{n}}e^{-a_{1}^{2}x_{1}^{2}}\cdots e^{-a_{n}^{2}x_{n}^{2}}dx_{1}\cdots dx_{n}\\ &=\int_{{\bf{R}}}\cdots\int_{\bf{R}}e^{-a_{1}^{2}x_{1}^{2}}dx_{1}\...
user284331's user avatar
  • 55.7k
15 votes

Apparent inconsistency of Lebesgue measure

One can explicitly demonstrate the failure of the argument for $\mathbb R_+$. Generalize to $\mathbb R$ as needed. Consider the following maps $f:\mathbb N_+^2 \rightarrow \mathbb N_+$ and $g: \...
eyeballfrog's user avatar
  • 22.8k
15 votes
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Understanding Layer Cake Representation

Here $\mathbb{1}_P(t)$ is defined to be $$\mathbb{1}_P(t) = \begin{cases} 1 & P(t) \text{ is true} \\ 0 & P(t) \text{ is false} \end{cases}$$ When $P$ is a set, like $[0, 3]$, then we look at $...
Chris Grossack's user avatar

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